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  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise

Transcript

Example 10 Find the term independent of x in the expansion of (3/2 ๐‘ฅ^2 " โˆ’ " 1/3๐‘ฅ)^6,x > 0. Calculating general term We know that general term of expansion (a + b)n is Tr + 1 = nCr (a)nโ€“r.(b)n For general term of expansion (3/2 ๐‘ฅ^2 " โˆ’ " 1/3๐‘ฅ)^6 Putting n = 6 , a = 3/2 ๐‘ฅ^2 , b = "โˆ’" 1/3๐‘ฅ Tr + 1 = 6Cr (๐Ÿ‘/๐Ÿ ๐’™^๐Ÿ )^(๐Ÿ” โˆ’ ๐’“) ((โˆ’๐Ÿ)/๐Ÿ‘๐’™)^๐’“ = 6Cr (3/2)^(6 โˆ’ ๐‘Ÿ) (๐‘ฅ^2 ")" ^((6 โˆ’ ๐‘Ÿ)) ((โˆ’1)/3 "ร—" 1/๐‘ฅ)^๐‘Ÿ = 6Cr (3/2)^(6 โˆ’ ๐‘Ÿ) (๐‘ฅ")" ^(2(6 โˆ’ ๐‘Ÿ)) ((โˆ’1)/3)^๐‘Ÿ (1/๐‘ฅ)^๐‘Ÿ = 6Cr (3/2)^(6 โˆ’ ๐‘Ÿ) (๐‘ฅ")" ^(12 โˆ’ 2๐‘Ÿ) ((โˆ’1)/3)^๐‘Ÿ (๐‘ฅ)^(โˆ’๐‘Ÿ) = 6Cr (3/2)^(6 โˆ’ ๐‘Ÿ) ((โˆ’1)/3)^๐‘Ÿ (๐‘ฅ")" ^(12 โˆ’ 2๐‘Ÿ) (๐‘ฅ)^(โˆ’๐‘Ÿ) = 6Cr (3/2)^(6 โˆ’ ๐‘Ÿ) ((โˆ’1)/3)^๐‘Ÿ (๐‘ฅ")" ^(12 โˆ’ 2๐‘Ÿ โˆ’ ๐‘Ÿ) = 6Cr (3/2)^(6 โˆ’ ๐‘Ÿ) ((โˆ’1)/3)^๐‘Ÿ (๐‘ฅ")" ^(12 โˆ’3๐‘Ÿ) We need to find the term independent of x So, power of x is 0 ๐‘ฅ^(12 โˆ’ 3๐‘Ÿ) = x0 Comparing powers 12 โ€“ 3r = 0 12 = 3r 12/3 = r 4 = r r = 4 Putting r = 4 in (1) T4+1 = 6C4 (3/2)^(6 โˆ’ 4) ((โˆ’1)/3)^4 (๐‘ฅ")" ^(12 โˆ’3(4)) T5 = 6C4 (3/2)^2 (1/3^4 ) (๐‘ฅ")" ^(12 โˆ’12) = 6C4 (3^2/2^2 )(1/3^4 ) (๐‘ฅ")" ^0 = 6C4 (1/2^2 )(3^2/3^4 ) (1) = 6C4 (1/2^2 )(1/3^2 ) = 6!/4!(6 โˆ’ 4)! (1/4)(1/9) = 6!/4!(2)! (1/4)(1/9) = (6(5)(4)!)/4!(2)! (1/4)(1/9) = (5 )/12 Hence, the term which is independent of x is 5th term = T5 = ๐Ÿ“/๐Ÿ๐Ÿ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.